Ieee Pes-ias Chapter 24-01-13

1/14/2013

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Transmission LinesElectricity's Highways

W.O. (Bill) Kennedy, P.Eng., FEIC

Principal

© 2013 b7kennedy & Associates Inc. 1

b7kennedy & Associates Inc.

Introduction

• Transmission lines are the highways for electricity.

• Their main purpose is to connect load to generation.

• Electricity in the context that I’ll use it includes both power and energy.

• Transmission lines are a civil, mechanical and electrical engineering challenge.

• This talk will focus on the electrical aspects of transmission lines.


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Transmission lines

• For the power system, there are all types of transmission lines from low voltage to high voltage.

• Each has its own purpose and each has a fixed amount of electricity transfer capability.

• Transmission lines are complex devices influenced by system voltages, loading, physical properties, system reactive capability and stability.

• We’ll examine each of these.• We will start with Surge Impedance Loading (SIL).• SIL is an important property of transmission lines.

SURGE IMPEDANCE LOADING


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Surge Impedance Loading (SIL)

• Transmission line consists of:– Shunt capacitance

– Series resistance and inductance

– Distributed along length of line

• Treat as distributed lumped elements

• Can ignore resistance



• Close the breaker at sending end

• Shunt capacitance charges to ½ CV2

• Close the breaker at receiving end and feed the load

• Series inductance consumes energy at ½ LI2

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Equating shunt and series energies

½ CV2 = ½ LI2

Performing the math yields

SIL (power) = V2/SI


System Model

• We need to develop a system model.• We will assume two machines at each end of our line.• Each machine will have sufficient reactive power

capability to supply and absorb the reactive power from/to the line.

• We know we can ‘collapse’ any power system to a two machine equivalent.

• Effectively, we’re developing a Thévenin voltage behind a Thévenin impedance.

• For the Thévenin impedance, we’ll use the breaker’s three phase fault rating.

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System Model

• There are three line models.• The first is the open circuit line. That is, how does a

transmission line open circuited at the receiving end behave? We’ll deal with this briefly.

• The second is the line with active and reactive power sources at both ends. This will be the subject to the seminar.

• The third is the line that has an active and reactive power source at the sending and a load at the receiving end.

• That is, the receiving end has no reactive power generation capability. We will not be discussing this type of line.

SIL Example


slack

0 MW

-106.75 Mvar

Send Bus Rec Bus

MW 0

0.00 Mvar

1.00 pu 0.00 Deg

1.03 pu-0.00 Deg

A

MVA

0.00 MW-106.75 Mvar

0.00 MW 0.00 Mvar

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Properties of Surge Impedance (SI)

• Remains fairly constant over a wide range of voltages.• Starts around 400 Ω at lower voltages and decreases

with bundling to around 225 Ω at 1500 kV.• Bundling – use of more than one conductor per phase.• Capacitance and inductance also remain fairly constant.• Using this, we can construct the following table.• Conductor size is expressed thousand circular mils or

MCM. MCM is an expression of conductor current carrying area.

• Capacitance or charging is expressed as kVAr/km.• R and X are in Ω/km or ohms/km.


Conductor Bundling

• Bundling is required at higher voltages.• Bundling is the use of two or more conductors per

phase.• Bundling reduces the series inductance and increases

the shunt capacitance.• The effect is to decrease SI and increase SIL.

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Properties of Transmission Lines


Voltage (kV)

Conductor (MCM)

SI (Ω) R (Ω/km) X (Ω/km)Charging (kVAr/km)

SIL (MW)

72 266 390 0.41 0.5 17 13

138 477 386 0.14 0.5 63 49

230 (single)

795 384 0.09 0.5 176 138

230 (bundled)

2x795 281 0.04 0.4 247 188

345 (bundled)

3x795 257 0.03 0.3 605 462

500 (bundled)

4x795 238 0.02 0.3 1,339 1051

ST. CLAIR CURVE


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St. Clair Curve

• Based on empirical knowledge of 1950’s.• Maximum loading at 50 miles (80 km) 3 x SIL – thermal

criteria.• Maximum loading at 300 miles (480 km) 1 x SIL.• St. Clair recognized relationship was not linear and drew

a curved line between the two points.• Assumed infinite VAr supply.• Reasonable for the sending end if a generator is present.• Not so reasonable for the receiving end if there’s no

reactive source.• We’ll assume that there’s sufficient reactive power at

both ends of the transmission line.


St. Clair Curve

0.0

0.5

1.0

1.5

2.0

2.5

3.0

50 150

250

350

450

550

650

750

850

950

Line Length in km

Lin

e L

oad

in S

IL

Normal Rating

Heavy Loading


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St. Clair Curve

• Dunlop et al in 1979 supplied analytical development for the St. Clair Curve.

• Up to 50 miles (80 km) loading based on thermal characteristics of the conductor – typically 3 x SIL.

• From 50 (80 km) miles to 200 (320 km) miles, loading dependent on voltage drop.

• Followed a 5% voltage drop criteria. Represents a heavily loaded transmission line.

• Above 200 miles (320 km) loading dependent upon Steady State Stability Limit.

• Set steady state stability limit to 30% (44o).


Dunlop et al.

• Used a two machine model.

• Assumed fully developed systems, i.e. high short circuit current or low short circuit impedance.

• Purpose was to extend the St. Clair Curve to UHV and line lengths approaching 1,000 km.

• Original St. Clair Curve went to 640 km.

• Set sending and receiving end Thévenin impedances equal to breaker rating (50 kA).

• Assumed infinite VAr supply, same as St. Clair.

• Results – confirmed validity of St. Clair Curve.


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500 kV Example

• Red dotted line – trend line.

• For this example, smooth changeover between criteria.

• 500 kV short circuit level = 50 kA.

500 kV Line Loading

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

2.00

2.25

2.50

2.75

3.00

3.25

010

020

030

040

050

060

070

080

090

010

00

Length (km)

Lin

e L

oad

ing

(S

IL)


230 kV Example

• Red dotted line – trend line

• Demonstrates change over from voltage to angle criteria

• 240 kV system short circuit value = 20 kA

240 kV Line Loading

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

2.00

2.25

2.50

2.75

3.00

3.25

010

020

030

040

050

060

070

080

090

010

00

Distance (km)

Lin

e L

oad

ing

(S

IL)


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Comparison

• Good correlation between curves.

• Lower short circuit levels mean lower loading capabilities.

• That is, line loading will lie below St. Clair Curve.

Line Loading

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

2.00

2.25

2.50

2.75

3.00

3.25

010

020

030

040

050

060

070

080

090

010

00

Length (km)

Lin

e L

oad

ing

(S

IL)

240 kV

500 kV

St. Claire


USING THE ST. CLAIR CURVE


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Using St. Clair

• Need to supply a 100 MW load 125 km from the nearest source.

• Assume a simple power system, that is a two machine equivalent.

• What supply voltage do we pick?

• From St. Clair we see that at 125 km, we can transfer 2.25 times SIL.

• From the table of SIL: 100 MW is twice SIL at 138 kV and approximately 75% of SIL at 230 kV.

• The question now becomes – what reactive support does the new load require?


REACTIVE POWER FLOW


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Reactive Support

• At SIL, shunt capacitive energy equals series inductive energy.

• Below SIL there is an excess of shunt capacitive energy.

• Above SIL there is a deficiency of shunt capacitive energy.

• Sending and receiving ends have to supply the reactive energy.


Calculating Reactive Support

• Use the per unit (pu) system.• Define SIL = 1 pu & Voltage = 1 pu.• At SIL, current = 1 pu.• At 1.5 SIL, current equals 1.5 pu.• Series reactive energy = ½ LI2

• Required reactive energy = 2.25 pu.• Line supplies 1 pu.• Therefore required reactive energy =1.25 pu• Voltages are equal, therefore ½ VArs from each end.


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345 kV, 200 km Line Example

• From the table, charging is 605 kVAr/km• VArs at 1.5 SIL are:

200 x 0.605 x 1.52 = 272 MVAr• VArs @ SIL = 120 MVAr• Required VArs from each end = (272 – 120)/2

= 76 MVAr• Two machine model has the capability• Receiving end may require shunt capacitor bank,

depends on system planning and design considerations for a single machine model.


CONDUCTOR SELECTION


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Conductor Selection

• REA specifies minimum conductor sizes.• Larger conductors at higher voltages required to

minimize corona which is rich in harmonics.• Electric field at conductor surface is minimized with a

large conductor – larger surface area.• Bundling, i.e. using more than one conductor per phase

at higher voltages reduces onset of corona.• Allows use of two smaller conductors.


CONDUCTOR RATINGS


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Conductor Ratings

• IEEE 738-2006 defines rating of conductors.• Based on modified House & Tuttle method.• Four ratings – Summer, Winter, Operating and Thermal.• Summer Thermal – conductor 100oC, ambient 40oC, wind 2 fps.• Winter Thermal – conductor 100oC, ambient 0oC, wind 2 fps.• Operating limits, based on 75oC conductor temperature with the

same ambient conditions.• Ambient temperatures are location dependent and utility operating

experience.• Wind speed is fixed!• Rating includes orientation of the line, that is does it run north –

south or east – west.• In addition, latitude of the line is taken into account.


Transmission Line Conductor Ratings


Voltage (kV)

Conductor (MCM)

Summer Thermal

MVA

Winter Thermal

MVA

Summer Operating

MVA

Winter Operating

MVA

72 266 45 70 60 80

138 477 125 190 165 215

230 (single)

795 280 435 385 500

230 (bundled)

2x795 560 875 765 1000

345 (bundled)

3x795 1260 1970 1725 2255

500 (bundled)

4x795 2430 3800 3330 4360

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LOSSES


Transmission Losses

• Because the X/R ratio for higher voltage lines is in the vicinity of 10, we have been able to ignore losses.

• However, losses are always there.

• Losses vary in relation to the line loading.

• Transmission lines have a long life – 35 to 50 years.

• It’s important that the right conductor be chosen because the associated losses are paid every year.

• In addition, line loading is increasing every year.


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Transmission Losses

Transmission Losses

0

100

200

300

400

500

4750 5000 5250 5500 5750 6000 6250 6500 6750 7000 7250 7500 7750

Net Generation to Supply Alberta Load (MW)

Lo

sses

(M

W)

• Losses are variable or stochastic.

• Simple system – losses vary as a square of current.

• Complex system –losses display a linear variance.


Transmission Losses

Transmission Losses Histogram

0

100

200

300

400

500

197

210

223

236

249

262

275

288

301

314

327

340

353

366

379

392

405

418

431

Losses (MW)

Co

un

t

• Histogram demonstrates a normal distribution pattern for losses.


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Losses on a Simple System

• Take a radial 138 kV line and load it up.

• Losses vary as the square of the current times the resistance of the line.

138 kV Radial Line Losses

0

1

2

3

4

5

6

7

12.5 25.0 37.5 50.0 62.5 75.0 87.5 100.0 112.5 125.0 137.5 150.0

Loading (MW)

Lo

sses

(M

W)


ECONOMIC CONDUCTOR SELECTION


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Conductor Optimization



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VOLTAGE AND ANGLE


Voltage and Angle

• We know that active power is closely coupled to the angle between voltage sources.

• For this reason we can transfer large amounts of active power over great distances.

• We also know that reactive power is closely coupled to the difference in voltage magnitudes.

• However, the differences in voltage magnitudes is limited to 5%.

• As a result it’s not possible to transmit large amounts of reactive power over the power system.

• Reactive should generated and consumed where it’s needed.


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Voltage and Angle


slack

Sending END 423 MW

41 Mvar

Receiving END

2 6%A

MVA

1.05 pu 1.00 pu

-410 MW

-14 Mvar 26%A

MVA

Gen BusLoad Bus

0.00 Deg -21.70 Deg

Active Power

Reactive Powerangle

voltage

Voltage and Angle


• Voltage drop measured between line terminals and limited to 5%.

• Angle measured across whole system and includes short circuit impedances at sending and receiving ends and limited to 44o

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Sending and Receiving Voltages

• Use Vr as reference

• Load is at unity power factor

• Vr and I are in phase

• 230 kV, 175 km line• Load is 1½ SIL = 282 MW• Vr = 230/√3 = 133 kV• I = 282/(√3 x 230) = 0.708 kA• X = 175 x 0.4 = 70 Ω• IX = 0.708 x 70 = 49.6 kV• Vs = √(1332 + 49.62) • Vs = 142 kV• δ = tan-1 (49.6/133) = 20.5o


Line Angle

• What is the angle across a line or system?

• How do you calculate it?

• Use the telegraph or long line equations.

• Find α + jβ

• Attenuation constant = α

• Phase constant = jβ

• Phase constant is in radians/km.

• Convert to δ or degrees.

• Want a simpler method.


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Line Angle

• Want a short cut method for calculating the angle across a system or section of a system.

• In a 60 Hz system, the system goes through 21,600 degrees in 1 second.

• Energy travels across the system at the speed of light.

• The speed of light is 3x105 km/sec.

• Therefore: system angle changes at 0.072 o/km.

• This is true at SIL of the system.

• Varies approximately in proportion to ratio of line loading to SIL.


Angle Across the Transmission Line

• Speed of light c = 3 x 105 km/sec.

• In one second on a 60 Hz system 21,600o

• Dividing c into the degrees gives the constant 0.072o/km.

• Proportional to SIL.• δ = 0.072 x 175 x 1½ =

19oVr

Vs

• Interested in calculating the angle between Vs & Vr.


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MAXIMUM POWER TRANSFER


• P = Re (Es* I)

• Note: correct way is: P = Re (Es I*)

Maximum Power Transfer


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• Steady State Stability Margin

• Typical value = 44o

• 70% power transfer

• Margin = 30%

• Calculated:

(100 x (1 – 0.7) / 1%


0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 20 40 60 80 100

120

140

160

180

Angle (degrees)

Po

wer

(p

u)


Stability – text book basics

• Po is the operating point.

• Typical value = 44o @ 70% power transfer.

• < 90o stable region.

• > 90o & < 135o transient region.

• > 135o unstable region.


0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 20 40 60 80 100

120

140

160

180

Angle (degrees)

Po

wer

(p

u)

Po

Stable

Transient

Unstable


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APPLICATION EXAMPLE


Application Example

• Let’s conclude with an application.

• You’re in the field, testing a 345 kV breaker.

• You’re finished for the day, and discover the 345 kV breaker won’t close.

• You trace the problem to the synchronizing check relay required for closing the breaker.

• What do you do?


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Ferranti Effect

• Open circuit voltage rise at the receiving end.• Can be worked out from first principles.• Travelling wave theory is best method.• Voltage rise equal to the inverse of the cosine of the

line angle.• Open end voltage rise on 230 kV, 250 km line is:


Synch Check

• Open 345 kV breaker won’t close.• Problem traced to Synch Check Relay – won’t

allow breaker to close.• Want to check the Synch Check Relay setting at

open breaker.


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Synch Check

• 345 kV parallel line 200 km in length each carrying SIL.

• With one line open, other line carries 2 SIL.• Angle is 29o.


Synch Check

• Design allows maximum voltage drop of 10%.

• Actual voltage drop is 12%.

• Open circuit voltage rise is 1.03 pu.

• Voltage difference across open breaker is 0.15 pu.


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Synch Check

• Expected relay setting would be minimum angular difference of 30o and minimum voltage difference of 15%.

• You discover relay is set at 40o and 10% voltage difference.

© 2013 b7kennedy & Associates Inc 59

Power of These Methods

• Assuming you become a power system engineer.

• Given a computer with up to $75k worth of very powerful software.

• Your supervisor will tell you – play with the system!

• Test it against NERC Criteria.

• How do know you’re getting the right answer?


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NERC Requirements

• Category A – all facilities in service, system must be stable.

• That is no voltage violations and no equipment overloaded.

• Category B – one piece of equipment out of service, system must be stable.

• Category C – two pieces out of service, system must be stable.

• Category D – three or more pieces out of service, load and or generator shedding, system must return to stable state.


Cautionary Note ...

• Understand these methods before you use them.

• Short cut methods are no substitute for detailed power flow and fault studies

• Can and should be used before you start your work!

• Can be used to validate your studies!• Develop your own numbers for the SIL table

as each power system is a little bit different.


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Transmission Line Summary

• Natural or SIL is an important property of transmission lines.

• SIL varies as the square of the line voltage.• St. Clair Curve shows loading is related to distance.• Overhead lines are most common for long distance

transmission.• Reactive power requirements vary as the loading of the

line.• Minimum conductor sizes dictated by corona.• Standards dictate maximum loading of conductor.• Transmission losses are fixed once conductor chosen.


That’s all folks!

• Discussion

• Questions


Documents

Ieee Pes-ias Chapter 24-01-13